贪心算法

Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d. 

We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates. 
 
Figure A Sample Input of Radar Installations

Input

The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases. 

The input is terminated by a line containing pair of zeros 

Output

For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.

 

 

判断需要建造最少的雷达时，需要将远的

 

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <math.h>

using namespace std;

struct point
{
    double left, right;
}p[2010], temp;

bool operator < (point a, point b)
{
    return a.left < b.left;
}

int main()
{
    int n;
    double r;
    int kase = 0;
    while (cin >> n >> r && (n || r))
    {
        bool flag = false;
        for (int i = 0; i < n; i++)
        {
            double a, b;
            cin >> a >> b;
            if (fabs(b) > r)
            {
                flag = true;
            }
            else
            {
                p[i].left = a * 1.0 - sqrt(r * r - b * b);
                p[i].right = a * 1.0 + sqrt(r * r - b * b);
            }
        }
        cout << "Case " << ++kase << ": ";
        if (flag)
        {
            cout << -1 << endl;
        }
        else
        {
            int countt = 1;
            sort(p, p + n);
            temp = p[0];

            for (int i = 1; i < n; i++)
            {
                if (p[i].left > temp.right)
                {
                    countt++;
                    temp = p[i];
                }
                else if (p[i].right < temp.right)
                {
                    temp = p[i];
                }
            }
            cout << countt << endl;
        }
    }
}

至今不懂